For the method of characteristics, parameterize by $x$; the characteristic curves satisfyThus $x^2+y=C$ and $u-\sin x$ is constant on each characteristic curve. The general form in the quadrant is $u=\sin x+F(x^2+y)$. The boundary condition gives $F(s^2)=\cos s-\sin s$ for every $s\ge 0$.

The characteristic curve through $(x,y)$ meets the boundary at $(s,0)$ with $s=\sqrt{x^2+y}\ge x$. Integrating backwards from that boundary value yieldsIndeed ${\partial }_x-2x{\partial }_y$ annihilates $x^2+y$ and sends $\sin x$ to $\cos x$, while on $y=0$ the expression reduces to $\cos x$. Every characteristic in the quadrant meets the prescribed boundary, so its value is determined. The displayed solution is smooth away from the origin. At the origin the boundary is characteristic and the solution is continuous but has a singular inward $y$ derivative; the differential equation is interpreted in the interior, with the boundary data attained continuously.

The fraction field of an integral domain $R$ is a field $F$ containing an identified copy of $R$ in which every element has the form $a/b$ with $a,b\in R$ and $b\ne 0$. Equality means $a/b=c/d$ exactly when $ad=bc$; the operations are the usual addition and multiplication of fractions. We use unital ring homomorphisms.

Since $\phi$ is injective, $\phi (b)\ne 0$ whenever $b\ne 0$. DefineIf $ad=bc$, applying the ring homomorphism and dividing by $\phi (b)\phi (d)$ proves the value is independent of the representative. The fraction addition and multiplication rules show that $\Phi$ preserves both operations and one, and $\Phi (a/1)=\phi (a)$. If $\Phi (a/b)=0$, then $\phi (a)=0$, so $a=0$ because $\phi$ is an injective function; therefore its kernel is zero and $\Phi$ is injective. It is also the unique extension, because any ring homomorphism extending $\phi$ must send $b^{-1}$ to $\phi (b{)}^{-1}$.

For a counterexample, take $R=\mathbb{Z}$, $F=\mathbb{Q}$, and reduction modulo two $\psi :\mathbb{Z}\to \mathbb{Z}/2\mathbb{Z}$. An extension would give $1=\Psi (2)\Psi (1/2)=0$, a contradiction. Nonzero denominators must become invertible in the target; an arbitrary ring homomorphism need not have that property.

Use completing the square rather than computing the eigenvalues of the symmetric matrix. Successively removing the mixed terms givesThus one suitable invertible linear change of coordinates isThe coordinate matrix is upper triangular with determinant one, so the change is an invertible linear map. Its inverse is $z=Z$, $y=Y+4Z/7$, $x=X-2Y+5Z/14$. Hence the diagonal coefficients can be taken as $2,-7,-3/14$. This is a matrix congruence of a quadratic form; an orthogonal transformation was not required.